$R/xR$ normal domain then $R$ is also a normal domain I'm not sure if this result is correct. I want to show "Say $R$ is a Noetherian local ring and $x$ is an $R$-regular element. Assume $R/xR$ is a normal domain. Then $R$ is also a normal domain."
"Domain" part proof: Say $yz=0$ where $y,z\neq 0$. Since $R$ is Noetherian, $y=x^ny'$ and $z=x^m z'$ where $y',z'\notin (x)$. Then $0=yz=x^{m+n}y'z'$. Since $x$ is a nonzerodivisor, it implies $y'z'=0$. Consider its image in $R/xR$, then $0$ is the product of two nonzero elements. (Contradiction)
So the "domain" property does deform. I cannot do the same thing for normality.
Attempt: Say $a/b\in Frac(R)$ is integral over $R$. Suppose we have the best case where $a,b\notin (x)$. Then we can consider the image of the equation in $R/xR$, then $\overline{a/b}$ is integral over $R/xR$, hence $a=rb+sx$. That does not tell me anything.
 A: I believe this result (at the stated level of generality) is due to Seydi [Seydi 1972, Proposition I.7.4] (see also [Seydi 1996, Corollaire 2.8]). We will instead follow the alternative proof from [Brezuleanu and Rotthaus 1982, Lemma 0].
Lemma. Let $(R,\mathfrak{m})$ be a Noetherian local ring and let $x \in \mathfrak{m}$ be a nonzerodivisor. If $R/xR$ is normal, then $R$ is normal.
Proof. By Serre's criterion for normality, it suffices to show that $R$ satisfies $R_1$ and $S_2$.
To show that $R$ satisfies $S_2$, we want to show that
$$
\operatorname{depth}(R_\mathfrak{p}) \ge \min\bigl\{2,\operatorname{height}(\mathfrak{p})\bigr\}
$$
for every prime ideal $\mathfrak{p} \subseteq R$. It suffices to consider $\mathfrak{p}$ for which $\operatorname{depth}(R_\mathfrak{p}) \le 2$. If $x \in \mathfrak{p}$, then $\operatorname{depth}((R/xR)_\mathfrak{p}) = \operatorname{depth}(R_\mathfrak{p})-1 < 2$. Since $R/xR$ is $S_2$, we have $\operatorname{height}(\mathfrak{p}\cdot(R/xR)) = \operatorname{depth}(R_\mathfrak{p})-1$, and hence
$$\operatorname{height}(\mathfrak{p}) = \operatorname{height}\bigl(\mathfrak{p}\cdot(R/xR)\bigr) + 1 = \operatorname{depth}(R_\mathfrak{p}).$$
On the other hand, if $x \notin \mathfrak{p}$, choose a prime ideal $\mathfrak{q}$ that is minimal over $\mathfrak{p}+xR$. Since $(\mathfrak{p}+xR)R_\mathfrak{q}$ is $\mathfrak{q}R_\mathfrak{q}$-primary, we then have
$$
\operatorname{depth}(R_\mathfrak{q}) = \operatorname{depth}\bigl((\mathfrak{p}+xR)R_\mathfrak{q},R_\mathfrak{q}\bigr) \le \operatorname{depth}(R_\mathfrak{p})+1.
$$
Then, we have $\operatorname{depth}((R/xR)_\mathfrak{q}) \le \operatorname{depth}(R_\mathfrak{p}) \le 2$, and since $R/xR$ is $S_2$, we have $\operatorname{height}(\mathfrak{q}\cdot(R/xR)) \le 2$, in which case
$\operatorname{height}(\mathfrak{q}) \le 3$. Finally, since $\mathfrak{q}$ is minimal over $\mathfrak{p} + xR$, we have $\operatorname{height}(\mathfrak{p}) \le 2$.
To show that $R$ satisfies $R_1$, we want to show that $R_\mathfrak{p}$ is regular for every prime ideal $\mathfrak{p} \subseteq R$ such that $\operatorname{height}(\mathfrak{p}) \le 1$. If $x \in \mathfrak{p}$, then $\mathfrak{p}\cdot(R/xR)$ is of height $0$. Since $R/xR$ satisfies $R_1$, we see that $(R/xR)_{\mathfrak{p}}$ is regular, which implies $R_\mathfrak{p}$ is regular. On the other hand, if $x \notin \mathfrak{p}$, choose a prime ideal $\mathfrak{q}$ that is minimal over $\mathfrak{p}+xR$. Since $\operatorname{height}(\mathfrak{q}) \ge 2$, we know that $\operatorname{depth}(R_\mathfrak{q}) \ge 2$ (since $R$ satisfies $S_2$ by the previous paragraph), and hence
$$\operatorname{height}\bigl(\mathfrak{q} \cdot (R/xR)\bigr) \ge \operatorname{depth}\bigl((R/xR)_\mathfrak{q}\bigr) \ge 1.$$
Since $R/xR$ satisfies $R_1$, we see that $(R/xR)_\mathfrak{q}$ is regular, and hence $R_\mathfrak{q}$ is regular. Since $R_\mathfrak{p}$ is a localization of $R_\mathfrak{q}$, we conclude that $R_\mathfrak{p}$ is also regular. $\blacksquare$
